Oscillation on Subset

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Theorem

Let $f: D \to \R$ be a real function where $D \subseteq \R$.

Let $x$ be a point in $D$.

Let $S_x$ be a set of real sets that contain (as an element) $x$.

Let $\map {\omega_f} I$ be the oscillation of $f$ on a set $I$ in $S_x$:

$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$


Let $I \in S_x$.

Let $\map {\omega_f} I \in \R$.

Let $J \in S_x$ be a subset of $I$.


Then:

$\map {\omega_f} J \in \R$
$\map {\omega_f} J \le \map {\omega_f} I$


Proof

Let:

$I, J \in S_x$
$J \subset I$
$\map {\omega_f} I \in \R$

where:

$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

We need to prove:

$\map {\omega_f} J \in \R$
$\map {\omega_f} J \le \map {\omega_f} I$


We intend to prove that $\map {\omega_f} J \in \R$.

We start by proving that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is bounded above and non-empty.


We have that $J$ is a subset of $I$.

Therefore:

$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a subset of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.

The statement $\map {\omega_f} I \in \R$ means that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ admits a supremum.

Therefore, $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above.

Accordingly:

$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is bounded above as $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a subset of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$


We observe that $\size {\map f y - \map f z} = 0$ for $y = z = x$.

Therefore, $0 \in \set {\size {\map f y - \map f z}: y, z \in J \cap D}$ as $x \in J \cap D$.

Accordingly:

$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is non-empty


We have that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a real set as $\size {\map f y - \map f z} \in \R$ for every $y, z \in D$.

We have shown that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is non-empty and bounded above.

Therefore, $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ admits a supremum by the Continuum Property.

In other words:

\(\displaystyle \sup \set {\size {\map f y - \map f z}: y, z \in J \cap D}\) \(\in\) \(\displaystyle \R\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \map {\omega_f} J\) \(\in\) \(\displaystyle \R\) definition


We finished proving that $\map {\omega_f} J \in \R$.

It remains to prove that $\map {\omega_f} J \le \map {\omega_f} I$.


We have:

$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a subset of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$
$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ admits a supremum
$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ admits a supremum

Then:

\(\displaystyle \sup \set {\size {\map f y - \map f z}: y, z \in J \cap D}\) \(\le\) \(\displaystyle \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}\) Supremum of Subset
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \map {\omega_f} J\) \(\le\) \(\displaystyle \map {\omega_f} I\) definitions

$\blacksquare$